Periodic(Trends( - Department of Chemistry · CEM 141 - Fall Semester 2012 – Exam 2 Rooms Monday,...
Transcript of Periodic(Trends( - Department of Chemistry · CEM 141 - Fall Semester 2012 – Exam 2 Rooms Monday,...
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Periodic Trends
• In this chapter we’ll explain why • We’ll then ra7onalize observed trends in
– Sizes of atoms and ions. – Ioniza7on energy. – Electron affinity.
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Effective Nuclear Charge • In a many-‐electron atom,
electrons are both aDracted to the nucleus and repelled by other electrons.
• The nuclear charge that an electron “feels” depends on both factors.
• It’s called Effec7ve nuclear charge.
• electrons in lower energy levels “shield” outer electrons from posi7ve charge of nucleus.
Na atom looks like this:
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Effec7ve Nuclear Charge
The effec7ve nuclear charge, Zeff, is:
Zeff = Z − S Where: Z = atomic number
S = screening constant, usually close to the number of inner (n-‐1) electrons.
Na
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• Example: Which element’s outer shell or “valence” electrons is predicted to have the largest Effec7ve nuclear charge? Kr, Cl or O?
Effec7ve Nuclear Charge
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Valence electrons Many chemical proper7es depend on the valence
electrons. Valence electrons: The outer electrons, that are
involved in bonding and most other chemical changes of elements.
Rules for defining valence electrons. 1. In outer most energy level (or levels) 2. For main group (representa7ve) elements (elements
in s world or p world) electrons in filled d or f shells are not valence electrons
3. For transi7on metals, electrons in full f shells or full d shells are not valence electrons.
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Examples: (valence electrons in blue) P: [Ne]3s23p3
As: [Ar] 4s23d104p3 I: [Kr]5s24d105p5 Ta: [Kr]6s24f145d3 Zn: [Ar]4s23d10
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Sizes of Atoms
The bonding atomic radius is defined as one-‐half of the distance between covalently bonded nuclei.
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Bonding atomic radius tends to… …decrease from lef to right across a row due to increasing Zeff.
…increase from top to boDom of a column due to increasing value of n
Sizes of Atoms
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Sizes of Ions
Ionic size depends upon:
Nuclear charge.
Number of electrons.
Orbitals in which electrons reside.
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Sizes of Ions
• Ca7ons are smaller than their parent atoms. – The outermost electron is removed and repulsions are reduced.
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Sizes of Ions • Anions are larger than their parent atoms. – Electrons are added and repulsions are increased.
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Sizes of Ions • Ions increase in size as you go down a column. – Due to increasing value of n.
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Sizes of Ions
• In an isoelectronic series, ions have the same number of electrons.
• Ionic size decreases with an increasing nuclear charge.
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atom/ion size examples
• Put the following in order of size, smallest to largest:
• Na, Na+, Mg, Mg2+, Al, Al3+, S, S2-‐, Cl, Cl-‐
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Atom size examples Al3+, Mg2+, Na+, Cl, S, Al, Mg, Na, Cl-‐, S2-‐
Start with atoms with no n=3 electrons, order isoelectronic by nuclear charge.
Next, neutral atoms highest Eff first
Last, anions, highest Eff first
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Ioniza7on Energy
• Amount of energy required to remove an electron from the ground state of a gaseous atom or ion. – First ioniza7on energy is that energy required to remove first electron.
– Second ioniza7on energy is that energy required to remove second electron, etc.
El -‐-‐-‐-‐-‐-‐-‐> El+ + e-‐ Na -‐-‐-‐-‐-‐-‐-‐> Na+ + e-‐
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Ioniza7on Energy
• It requires more energy to remove each successive electron.
• When all valence electrons have been removed, the ioniza7on energy takes a quantum leap.
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Trends in First Ioniza7on Energies • going down a
column, less energy to remove the first electron. – For atoms in the
same group, Zeff is essen7ally the same, but the valence electrons are farther from the nucleus.
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Trends in First Ioniza7on Energies • Generally, it gets harder to remove an electron going across. – As you go from lef to to right, Zeff increases.
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Trends in First Ioniza7on Energies On a smaller scale, there are two jags in each line. Why?
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Trends in First Ioniza7on Energies
• The first occurs between Groups IIA and IIIA.
• Electron removed from p-‐orbital rather than s-‐orbital – Electron farther from nucleus
– Small amount of repulsion by s electrons.
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Trends in First Ioniza7on Energies • The second occurs between Groups VA and VIA. – Electron removed comes from doubly occupied orbital.
– Repulsion from other electron in orbital helps in its removal.
versus:
N O
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Electron Affinity
Energy change accompanying addi7on of electron to gaseous atom:
Cl + e− ⎯⎯→ Cl−
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Trends in Electron Affinity
In general, electron affinity becomes more exothermic as you go from lef to right across a row.
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Trends in Electron Affinity
There are also two discon7nui7es in this trend.
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Trends in Electron Affinity
• The first occurs between Groups IA and IIA. – Added electron must go in p-‐orbital, not s-‐orbital.
– Electron is farther from nucleus and feels repulsion from s-‐electrons.
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Trends in Electron Affinity
• The second occurs between Groups IVA and VA. – Group VA has no empty orbitals.
– Extra electron must go into occupied orbital, crea7ng repulsion.
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Chemical Bonds Three types:
– Ionic Electrostatic attraction
between ions
Covalent Sharing of electrons
Metallic Metal atoms bonded to several other atoms
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Bonding a con7nuum
Cs
Na3Bi
Na3Sb
K2Se KCl
CsF AlF3 PF5 ClF F2
I2
S
As
Sn
Metallic
Covalent Ionic
Both electroposi7ve
Both electronega7ve
One electroposi7ve One electronega7ve
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Ionic Bonding
When a metal and a nonmetal get together
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Energetics of Ionic Bonding
it takes 495 kJ/mol to remove 1 electron from sodium.
495x2 = 990 kJ/2 Na
2Na(s) + Cl2(g) -------> 2NaCl(s)
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Energetics of Ionic Bonding
We get 349 kJ/mol Cl back by giving 1 electron to each 1 mole of Cl2.
-349x2 = -700 kJ/mol Cl2
990 kJ/2Na – 700 kJ/Mol Cl2 = 290kJ
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Energetics of Ionic Bonding
• But these numbers don’t explain why the reaction of sodium metal and chlorine gas to form sodium chloride is so exothermic!
990 kJ/2Na – 700 kJ/Mol Cl2 = 290kJ
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Energetics of Ionic Bonding • There must be a
third piece to the puzzle.
• What is as yet unaccounted for is the electrostatic attraction between the newly formed sodium cation and chloride anion.
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Lattice Energy
• This third piece of the puzzle is the lattice energy: The energy required to completely separate a mole of
a solid ionic compound into its gaseous ions. • The energy associated with electrostatic
interactions is governed by Coulomb’s law:
Eel = κ Q1Q2 d
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Lattice Energy
Eel = κ Q1Q2 d
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Lattice Energy • Lattice energy, then, increases with the charge on
the ions. • It also increases with decreasing size of ions.
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Energetics of Ionic Bonding
By accounting for all three energies (ionization energy, electron affinity, and lattice energy), we can get a good idea of the energetics involved in such a process.
Na(s) + 1/2Cl2(g) -----> NaCl(s)
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CEM 141 - Fall Semester 2012 – Exam 2 Rooms Monday, October 22nd
7:15pm-8:15pm
Section Room TA
1,2,3,4,5,6,7,8,9,10 N130 Business Complex Koyeli, Mike, Robert
11,12,13,14,15,16,17,18,19,20 B115 Wells Hall Yuan, Ben, Wenjing
21,22,23,24,25,26,27 138 Chemistry Ruiqiong, Denise, Zhongqi
28,29,30,31 118 Psychology Bldg. Ryan, Penghao
32,33,34,35 B117 Wells Hall Xiaoxiao, Lihui
36,37,38,75,76 N100 Business Complex Chenjia, Shuai, Scott,
Mersedeh
39,40,41 128 Nat. Sci. Bldg. Sujana, Shannon
42,43,44,45 1345 Engineering Bldg. Jerome, Shuang
46,47,48,49,50,51,52,53 102 Conrad Hall Preston, Nick, Krystin
54,55,56,57,58,59,60,61,62,63 1281 Anthony Hall Ruipeng, Zhefei, Adeayo, Punsisi
64,65,66,67 402 Computer Center Lindsey, Xiaoran
68,69,70 1279 Anthony Hall Christopher, Shuxuan
71,72,73,74 N101 North Kedzie Ali, Zahra
77,78,79,80 B119 Wells Hall Chen, Zhiling
81,82,83,84 1410 BioMed & Phys. Sci Monica, Brandon
Alternate Exam Monday, October 22nd 6:45am-7:45am room 138 Chemistry: Yan, Chengpeng
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Energetics of Ionic Bonding
• These phenomena also help explain the “octet rule.”
• Elements tend to lose or gain electrons once they aDain a noble gas configura7on because energy would be expended that cannot be overcome by latce energies.
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Covalent Bonding
• In these bonds atoms share electrons.
What happens when nonmetals get together
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Covalent Bonding
• There are several electrostatic interactions in these bonds: – Attractions between electrons
and nuclei – Repulsions between electrons – Repulsions between nuclei
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Polar Covalent Bonds • Although atoms often
form compounds by sharing electrons, the electrons are not always shared equally.
• Fluorine pulls harder on the shared electrons than hydrogen does.
• Therefore, the fluorine end has more electron density than the hydrogen end.
• But how do you know who pulls hardest?
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Electronegativity: Developed 1st by Linus Pauling like this:
H-F -à H + F >> H-H -à H + H or F-F à F + F Why? Because there is an ionic component to attraction in H-F F more – and H more + so the ionic component makes bond stronger.
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Electronegativity:
• The ability of atoms in a molecule to attract electrons to itself.
• On the periodic table, electronegativity increases as you go… – …from left to right across a row. – …from the bottom to the top of a column.
A measure of how much an atom attracts electrons when it is in a molecule.
Developed 1st by Linus Pauling like this:
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Electronegativity:
• On the periodic chart, electronegativity increases as you go… – …from left to right across a row. – …from the bottom to the top of a column.
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Polar Covalent Bonds • When two atoms share
electrons unequally, a bond dipole results.
• The dipole moment, µ, produced by two equal but opposite charges separated by a distance, r, is calculated:
µ = Qr • It is measured in debyes (D).
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Polar Covalent Bonds
The greater the difference in electronegativity, the more polar is the bond.
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Covalent bonds
Interatomic distance
E
Repulsion between nuclei
Distance between nuclei
Bond energy
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Lewis Structures
Lewis structures are representations of molecules showing all valence electrons, bonding and nonbonding.
Lines correspond to 2 electrons in bond
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Writing Lewis Structures
1. Find the sum of valence electrons of all atoms in the polyatomic ion or molecule. – If it is an anion, add one
electron for each negative charge.
– If it is a cation, subtract one electron for each positive charge.
PCl3
5 + 3(7) = 26
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Writing Lewis Structures 2. The central atom is
the least electronegative element that isn’t hydrogen (why?). Connect the outer atoms to it by single bonds.
Keep track of the electrons: 26 -‐ 6 = 20
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Writing Lewis Structures
3. Fill the octets of the outer atoms.
Keep track of the electrons: 26 -‐ 6 = 20 -‐ 18 = 2
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Writing Lewis Structures
4. Fill the octet of the central atom.
Keep track of the electrons: 26 -‐ 6 = 20 -‐ 18 = 2 -‐ 2 = 0
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Writing Lewis Structures
5. If you run out of electrons before the central atom has an octet…
…form multiple bonds until it does.
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Writing Lewis Structures • Then assign formal charges.
– For each atom, count the electrons in lone pairs and half the electrons it shares with other atoms.
– Subtract that from the number of valence electrons for that atom: The difference is its formal charge.
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Writing Lewis Structures
• The best Lewis structure… – …is the one with the fewest charges. – …puts a negative charge on the most
electronegative atom.
-2 0 +1 -1 0 0 0 0 -1
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Resonance
Draw the Lewis structure for ozone, O3.
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Resonance
Draw the Lewis structure for ozone, O3.
But why should one O be different from the other?
-‐
+
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Resonance
• It is at odds with the true, observed structure of ozone, – …both O—O bonds
are the same length. – …both outer
oxygens have a charge of -1/2.
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Resonance
• One Lewis structure cannot accurately depict a molecule such as ozone.
• We use multiple structures, resonance structures, to describe the molecule.
+ -‐
-‐ +
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Resonance Just as green is a synthesis of
blue and yellow…
…ozone is a synthesis of these two resonance structures.
It is not jumping between the two.
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Resonance Draw resonance structure for: HCO2
-
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Resonance Draw Lewis structure for: HCO2
-
O C O!! H!
..
.. .. .. ..
-
But why would the two oxygens be different?
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Resonance • In truth the electrons that make up the double
bond are not localized, but rather are delocalized.
O C O!! H!
..
.. .. .. ..
- O C O!! H!
..
.. .. .. ..
-
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Resonance • Draw the Lewis structure of NO3
-
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Resonance • Draw the Lewis structure of NO3
-
N
O
O O .. .. ..
.. .. ..
.. -
+ N
O
O O .. .. ..
..
.. ..
.. - + N
O
O O .. .. ..
..
.. ..
.. - +
.. .. .. - -
..
-
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Resonance • The organic compound
benzene, C6H6 is a hexagon of carbon atoms with 6 H/s Draw the Lewis structure for benzene.
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Resonance • The organic compound
benzene, C6H6, has two resonance structures.
• It is commonly depicted as a hexagon with a circle inside to signify the delocalized electrons in the ring.
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Exceptions to the Octet Rule
• There are three types of ions or molecules that do not follow the octet rule: – Ions or molecules with an odd number of
electrons. – Ions or molecules with less than an octet. – Ions or molecules with more than eight
valence electrons (an expanded octet).
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Odd Number of Electrons
Though relatively rare and usually quite unstable and reactive, there are ions and molecules with an odd number of electrons.
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Odd Number of Electrons • Example: NO
N O .. ..
.. .
What’s nitric oxide good for?
N O .. .. . - + ..
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Fewer Than Eight Electrons
Draw the Lewis structure for BF3:
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Fewer Than Eight Electrons
Draw the Lewis structure for BF3:
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Fewer Than Eight Electrons
• Consider BF3: – Giving boron a filled octet places a negative
charge on the boron and a positive charge on fluorine.
– This would not be an accurate picture of the distribution of electrons in BF3.
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Fewer Than Eight Electrons Therefore, structures that put a double bond between boron and fluorine are much less important than the one that leaves boron with only 6 valence electrons.
Double bonds to halogens, especially F don’t happen.
+
- - - + +
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Fewer Than Eight Electrons The lesson is: If filling the octet of the central atom results in a negative charge on the central atom and a positive charge on the more electronegative outer atom, don’t fill the octet of the central atom.
+
+ + - - -
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More Than Eight Electrons
Draw the Lewis structure for PCl5
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More Than Eight Electrons • The only way PCl5 can
exist is if phosphorus has 10 electrons around it.
• atoms on the 3rd row or below can go over an octet of electrons – Presumably d orbitals in
these atoms participate in bonding.
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More Than Eight Electrons
• Draw the Lewis structure for phosphate • PO4
-3
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More Than Eight Electrons Even though we can draw a Lewis structure for the phosphate ion that has only 8 electrons around the central phosphorus, a common Lewis structure puts a double bond between the phosphorus and one of the oxygens.
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More Than Eight Electrons • This eliminates the charge on the phosphorus
and the charge on one of the oxygens. • The lesson is: When the central atom is on the
3rd row or below and expanding its octet eliminates some formal charges, you can do so.
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More Practice • Draw lewis structures for: • SO4
-2, CO3-2, CHCl3, CN3H6
+ (H’s are attached to the N’s). SO2, PO3
2-, NO2, BrO3-,
• ClO4-,
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..
SO4-‐2, CO3
-‐2, CHCl3, CN3H6+ (H’s on N’s). SO2, PO3
1-‐, NO, BrO3-‐, ClO4
-‐,
S
O
O
O
O
..
.. .. .. ..
..
: :
: :
: : - -
-
-
+2 C
O : :
O ..
: : - O ..
: : - C Cl
.. :
.. Cl : .. Cl : ..
:
H
C
N H H
N H
H
.. N ..
H H
+
S O O .. .. .. : : - .. + P
O : :
O ..
: : - O ..
: : - +
N O .. .. : .
Cl
O
O
O
O
..
.. .. .. ..
..
: :
: :
: : - -
-
-
+3
Br
O : :
O ..
: : - O ..
: : .
+2 -
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Covalent Bond Strength
• The strength of a bond is measured by determining how much energy is required to break the bond.
• This is the bond enthalpy. • The bond enthalpy for a Cl—Cl bond,
D(Cl—Cl), is 242 kJ/mol.
ΔH = 242 kJ/mol
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Average Bond Enthalpies
• Average bond enthalpies are positive, because bond breaking is an endothermic process.
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Average Bond Enthalpies
NOTE: These are average bond enthalpies, not absolute bond enthalpies; the C—H bonds in methane, CH4, will be a bit different than the C—H bond in chloroform, CHCl3.
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Enthalpies of Reaction
• Can use bond enthalpies to es7mate ΔH for a reac7on ΔHrxn = Σ(bond enthalpies of
bonds broken) -‐ Σ(bond enthalpies of bonds
formed)
This is a fundamental idea in chemical reactions. The heat of a reaction comes from breaking bonds and remaking bonds.
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Enthalpies of Reaction
CH4(g) + Cl2(g) ⎯⎯→ CH3Cl(g) + HCl(g)
In this example, one C—H bond and one Cl—Cl bond are broken; one C—Cl and one H—Cl bond are formed.
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Enthalpies of Reaction
So, ΔHrxn = [D(C—H) + D(Cl—Cl) - [D(C—Cl) + D(H—Cl)
= [(413 kJ) + (242 kJ)] - [(328 kJ) + (431 kJ)] = (655 kJ) - (759 kJ) = -104 kJ
CH4(g) + Cl2(g) ⎯⎯→ CH3Cl(g) + HCl(g)
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Bond Enthalpy and Bond Length
• We can also measure an average bond length for different bond types.
• As the number of bonds between two atoms increases, the bond length decreases.
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Enthalpy problem: • Calculate the enthalpy for: • CH4 + O2 ---> CO2 + H2O
• HC CH + O2 ------> CO2 + H2O
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Enthalpy problem: • Calculate the enthalpy for: • CH4 + 3/2O2 ---> CO2 + 2H2O • 4(C--H) + 3/2(O==O) - 2(C==O) - 4(OH) • 4(413) + 3/2(495) - 2(800) - 4(463) = -563 kJ
• HC CH + 5/2O2 ------> 2CO2 + H2O
• 1(CC) + 2CH + 5/2(O=O) - 4(C==O) - 2(OH) • 1(834) + 2(413) +5/2(495) - 4(800) - 2(463) = -1229 kJ